Evaluate the integrals in Exercises 53–58.
∫ sin(2x) cos...

Question

Evaluate the integrals in Exercises 53–58.

∫ sin(2x) cos(3x) dx

Accepted Answer

Welcome back everyone. Evaluate the the integral sine 5 x, cosine 2 x dx. For this problem we're going to rewrite the product in terms of a sum. So to begin with, let's remember that if we have sine of alpha multiplied by cosine of beta, we can rewrite it as 1/2 in parent sine of alpha plus beta plus sine of alpha minus beta. And in this case, Alpha is 5 X. And beta is 2x, so we get 1/2. Sign of 5 x plus 2 x gives us 7 x plus sin of 5 x minus 2x gives us 3 x. So essentially what we want to do is integrate. One half we can factor it out because it's a constant. Sign of 7 x. Plus sign of 3 XDX. So now we have elementary integrals. First of all, we can write one half in, begin by integrating sin of 7 x that be. Negative cosine, right, because the integral of sine x is negative cosine x. So in this case we would have negative cosine of 7 x divided by 7 according to the reverse chain rule. Since this is a composite function, we're dividing by the slope of the linear term which is 7, and similarly, the integral of sine of 3 x is negative cosine of 3 x divided by 3. And we're going to add a constant of integration C. Finally, let's distribute to get -1 divided by 14. Cosine of 7 x minus 1 divided by 6. Cosine of 3 x plus c. That's our final answer, and thank you for watching.

Evaluate the integrals in Exercises 53–58.∫ sin(2x) cos(3x) dx

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Evaluate the integrals in Exercises 53–58.
∫ sin(2x) cos(3x) dx